login
a(n) = a(n-1) + 3*a(n-2) -2*a(n-3) - 2*a(n-4), where a(0) = -1, a(1) = 2, a(2) = 3, a(3) = 4.
1

%I #6 Aug 27 2021 21:20:50

%S -1,2,3,4,11,13,32,41,89,122,243,349,656,973,1757,2666,4679,7217,

%T 12408,19369,32801,51658,86507,137141,227744,362837,598773,957514,

%U 1572671,2521993,4127432,6633041,10826009,17426282,28383363,45744109,74389616,120002653

%N a(n) = a(n-1) + 3*a(n-2) -2*a(n-3) - 2*a(n-4), where a(0) = -1, a(1) = 2, a(2) = 3, a(3) = 4.

%C a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622), so that a( ) has the growth rate of the Fibonacci numbers (A000045).

%H Clark Kimberling, <a href="/A295721/b295721.txt">Table of n, a(n) for n = 0..2000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (1, 3, -2, -2)

%F a(n) = a(n-1) + a(n-3) + a(n-4), where a(0) = -1, a(1) = 2, a(2) = 3, a(3) = 4.

%F G.f.: (-1 + 3 x + 4 x^2 - 7 x^3)/(1 - x - 3 x^2 + 2 x^3 + 2 x^4).

%t LinearRecurrence[{1, 3, -2, -2}, {-1, 2, 3, 4}, 100]

%Y Cf. A001622, A000045, A005672.

%K easy,sign

%O 0,2

%A _Clark Kimberling_, Nov 29 2017